Some ways linear algebra is different in infinite dimensions

There’s no notion of continuity in linear algebra per se. It’s not part of the definition of a vector space. But a finite dimensional vector space over the reals is isomorphic to a Euclidean space of the same dimension, and so we usually think of such spaces as Euclidean. (We’ll only going to consider real vector spaces in this post.) And there we have a notion of distance, a norm, and hence a topology and a way to say whether a function is continuous.

Continuity

In finite dimensional Euclidean space, linear functions are continuous. You can put a different norm on a Euclidean space than the one it naturally comes with, but all norms give rise to the same topology and hence the same continuous functions. (This is useful in numerical analysis where you’d like to look at a variety of norms. The norms give different analytical results, but they’re all topologically equivalent.)

In an infinite dimensional normed space, linear functions are not necessarily continuous. If the dimension of a space is only a trillion, all linear functions are continuous, but when you jump from high dimension to infinite dimension, you can have discontinuous linear functions. But if you look at this more carefully, there isn’t a really sudden change.

If a linear function is discontinuous, its finite dimensional approximations are continuous, but the degree of continuity is degrading as dimension increases. For example, suppose a linear function stretches the nth basis vector by a factor of n. The bigger n gets, the more the function stretches in the nth dimension. As long as n is bounded, this is continuous, but in a sense it is less continuous as n increases. The fact that the infinite dimensional version is discontinuous tells you that the finite dimensional versions, while technically continuous, scale poorly with dimension. (See practical continuity for more discussion along these lines.)

Completeness

A Banach space is a complete normed linear space. Finite dimensional normed spaces are always complete (i.e. every sequence in the space converges to a point in the space) but this might not happen in infinite dimensions.

Duals and double duals

In basic linear algebra, the dual of a vector space V is the space of linear functionals on V, i.e. the set of linear maps from V to the reals. This space is denoted V*. If V has dimension n, V* has dimension n, and all n-dimensional spaces are isomorphic, so the distinction between a space and its dual seems pedantic. But in general a Banach space and its dual are not isomorphic and so its easier to tell them apart.

The second dual of a vector space, V** is the dual of the dual space. In finite dimensional spaces, V** is naturally isomorphic to V. In Banach spaces, V is isomorphic to a subset of V**. And even when V is isomorphic to V**, it might not be naturally isomorphic to V**. (Here “natural” means natural in the category theory sense of natural transformations.)

I’m surprised that the V V** isomorphism (when it exists) isn’t natural. My intuition about naturality is that it happens when there is one transformation that stands out as more obvious than any others, it’s likely to be natural. And the embedding of V into V** by point evaluation really does stand out.

Often it seemed like a particular category left off some structure that my intuition wouldn’t ignore, and that made statements about that category seem non-intuitive. I wonder if there is some richer category than “Vector Spaces” in which V V** is natural.

V is never _naturally_ isomorphic to V**. Given an element of V there is no natural element corresponding in V** (just try to construct one). Both spaces have the same dimension but there is no _natural_ correspondence. (I mention the metric because for finite dimensional spaces a particular metric –there are many– is a particular isomorphism between V and V**). One thing that is confusing is that you could say that the corresponding element of the dual of the vector with coordinates (1, 0, 0) is [1, 0, 0] (“column vector”). But that already implies a certain basis (and in other basis the correspondence is not “natural”).